Question

Why is AM=MB=MC in this right triangle?

I understand that AM=BC because M is the centerpoint, but I do not understand MC.

Answer 1

Please refer to the Explanation.

Explanation:

`M` is the midpoint of `AB` and `/_ACB=90^@`.

Therefore, `C` lies on the (semi)circle described on `AB,` taking `AB`

as diameter.

`M` is the midpoint of the diameter `AB`.

`:. M` is the centre of the circle.

Thus, `C` lies on the circle having centre at `M` and a diameter `AB`.

Clearly, `MA=MB=MC"(=the radius of the circle)"`.

Answer 2

Please see below.

Explanation:

Your Question :
I understand that `AM=color(red)(BC)` because `M` is

the centerpoint, but I do not understand `MC.`

I think it is `AM=MB and` not `color(red)(BC)`

We have two Theorem :

`color(green)((1) "An angle inscribed in a semicircle is a right angle"`

`color(green)((2)"The circle whose diameter is the hypotenuse of the "`

`color(green)("right triangle , passes through three vertices of the triangle."`

So, `A ,B, and C` are the points on the circle and `mangleC=90^circ`

`:.color(violet)( " Hypotenuse AB of triangle = Diameter AB of circle"`

Now , `M " is the midpoint of AB"`

So, `"M is the center of the circle. "`

Let `color(blue)(r)` be the radius of circle.

`:.color(blue)(AM= MB=r`

Here, `bar(MC)` is the line segment joining center `M` and vertex `C`

`:. MC " is the radius of circle."`

`:.color(blue)( MC =r)`

Hence ,

`color(blue)(AM=MB=MC=r`